A wheel with 8 metallic spokes each 50 cm long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of the earth's magnetic field. The earth's magnetic field at the place is 0.4 G and the angle of dip is 60°. The induced emf between the axle and the rim of the wheel is:

$3.14×10^{-5} V$

The correct answer is Option (3) → $3.14×10^{-5} V$

The induced emf between the axle and the rim of a wheel rotating in a magnetic field is given by,

$ε= \frac{1}{2}B_h\omega R^2$

Horizontal component,

$B_H = B\cos 60°$

$=0.4×10^{-4}×\frac{1}{2}$

$=0.2×10^{-4}T$

Angular velocity, $ω=120×\frac{2\pi}{60}=4\pi\,rad/s$

$∴ε= \frac{1}{2}×0.2×10^{-4}×4\pi×(0.5)^2$

$=0.1×10^{-4}×\pi$

$≈3.14×10^{-5} V$